4-51
SECTION 4.6
.
.
Integration by Substitution
393
2.
When solving differential equations of the form
dy
dt
=
f
(
y
)
for the unknown function
y
(
t
), it is often convenient to make
use of a
potential function
V
(
y
). This is a function such that
−
dV
dy
=
f
(
y
). For the function
f
(
y
)
=
y
−
y
3
, find a potential
function
V
(
y
). Find the locations of the local minima of
V
(
y
)
and use a graph of
V
(
y
) to explain why this is called a “double-
well” potential. Explain each step in the calculation
dV
dt
=
dV
dy
dy
dt
= −
f
(
y
)
f
(
y
)
≤
0
.
Since
dV
dt
≤
0, does the function
V
increase or decrease as
time goes on? Use your graph of
V
to predict the possible val-
ues of lim
t
→∞
y
(
t
). Thus, you can predict the limiting value of
the solution of the differential equation without ever solving
the equation itself. Use this technique to predict lim
t
→∞
y
(
t
) if
y
=
2
−
2
y
.
3.
Let
f
n
(
x
)
=
2
n
+
4
n
2
x
−
1
2
n
≤
x
≤
0
2
n
−
4
n
2
x
0
≤
x
≤
1
2
n
0
otherwise
for
n
=
1
,
2
,
3
,
....
For an arbitrary
n
, sketch
y
=
f
n
(
x
)
and show that
1
−
1
f
n
(
x
)
dx
=
1. Compute lim
n
→∞
1
−
1
f
n
(
x
)
dx
.
For
an
arbitrary
x
=
0
in
[
−
2
,
2],
compute
lim
n
→∞
f
n
(
x
)
and
compute
1
−
1
lim
n
→∞
f
n
(
x
)
dx
.
Is
it
always
true
that
lim
n
→∞
1
−
1
f
n
(
x
)
dx
=
1
−
1
lim
n
→∞
f
n
(
x
)
dx
?
4.6
INTEGRATION BY SUBSTITUTION
In this section, we expand our ability to compute antiderivatives by developing a useful
technique called
integration by substitution.
This method gives us a process for helping
to recognize a whole range of new antiderivatives.
EXAMPLE 6.1
Finding an Antiderivative by Trial and Error
Evaluate
2
xe
x
2
dx
.
Solution
We need to find a function
F
(
x
) for which
F
(
x
)
=
2
xe
x
2
. You might be
tempted to guess that since
x
2
is an antiderivative of 2
x
,
F
(
x
)
=
x
2
e
x
2
is an antiderivative of 2
xe
x
2
. To see that this is incorrect, observe that, from the product
rule,
d
dx
(
x
2
e
x
2
)
=
2
xe
x
2
+
x
2
e
x
2
(2
x
)
=
2
xe
x
2
.
So much for our guess. Before making another guess, look closely at the integrand.
Notice that 2
x
is the derivative of
x
2
and
x
2
already appears in the integrand, as the
exponent of
e
x
2
. Further, by the chain rule, for
F
(
x
)
=
e
x
2
,
F
(
x
)
=
e
x
2
d
dx
(
x
2
)
=
2
xe
x
2
,
which is the integrand. To finish this example, recall that we need to add an arbitrary
constant, to get
2
xe
x
2
dx
=
e
x
2
+
c
.
More generally, recognize that when one factor in an integrand is the derivative of
another part of the integrand, you may be looking at a chain rule derivative.
394
CHAPTER 4
.
.
Integration
4-52
Note that, in general, if
F
is any antiderivative of
f
, then from the chain rule, we
have
d
dx
[
F
(
u
)]
=
F
(
u
)
du
dx
=
f
(
u
)
du
dx
.
From this, we have that
f
(
u
)
du
dx
dx
=
d
dx
[
F
(
u
)]
dx
=
F
(
u
)
+
c
=
f
(
u
)
du
,
(6.1)
since
F
is an antiderivative of
f
. If you read the expressions on the far left and the far right
sides of (6.1), this suggests that
du
=
du
dx
dx
.
